Question 31. If f(x)=\begin{cases}\frac{2^{x+2}-16}{4^x-16},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases} is continuous at x = 2, find k.
Solution:
Given that,
f(x)=\begin{cases}\frac{2^{x+2}-16}{4^x-16},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases} Also, f(x) is continuous at x = 2
So, LHL = RHL = f(2) .....(i)
Now,
f(2) = k ......(ii)
Let us consider LHL,
\lim_{x\to2^-}f(x) =\lim_{h\to0}f(2-h)
=\lim_{h\to0}\frac{2^{(2-h)+2}-16}{4^{(2-h)}-16}
=\lim_{h\to0}\frac{2^{4-h}-16}{4^{(2-h)}-16}
=\lim_{h\to0}\frac{2^4.2^{-h}-16}{4^2.4^{-h}-16}
=\lim_{h\to0}\frac{16.2^{-h}-16}{16.4^{-h}-16}
=\lim_{h\to0}\frac{16(2^{-h}-1)}{16(4^{-h}-1)}
=\lim_{h\to0}\frac{2^{-h}-1}{(2^{-h})^2-1^2}
=\lim_{h\to0}\frac{2^{-h}-1}{(2^{-h}-1)(2^{-h}+1)}=1/2 ......(iii)Using eq(i), (ii) and (iii), we get
k = 1/2
Question 32. If f(x)=\begin{cases}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases} is continuous at x = 0, find k.
Solution:
Given that,
f(x)=\begin{cases}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases} Also, f(x) is continuous at x = 2
So, LHL = RHL
Now,
\lim_{x\to0}f(x)=f(0) ⇒
\lim_{x\to0}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1}=k ⇒
\lim_{x\to0}\frac{1-sin^2x-sin^2x-1}{\sqrt{x^2+1}-1}=k ⇒
\lim_{x\to0}\frac{-2sin^2x}{\sqrt{x^2+1}-1}=k ⇒
\lim_{x\to0}\frac{-2(sin^2x)(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)}=k ⇒
\lim_{x\to0}\frac{-2(sin^2x)(\sqrt{x^2+1}+1)}{x^2}=k ⇒
-2\lim_{x\to0}\frac{(sin^2x)(\sqrt{x^2+1}+1)}{x^2}=k ⇒
-2\lim_{x\to0}(\frac{sinx}{x})^2\lim_{x\to0}(\sqrt{x^2+1}+1)=k ⇒ -2 × 1 × (1 + 1) = k
⇒ k = -4
Question 33. Extend the definition of the following by continuity f(x) = \frac{1-cos7(x-π)}{5(x-π)^2} at the point x = π.
Solution:
Given that,
\frac{1-cos7(x-π)}{5(x-π)^2} As we know that a f(x) is continuous at x = π if,
LHL = RHL = f(π) ......(i)
Let us consider LHL,
\lim_{x\toπ^-}f(x) =\lim_{h\to0}f(π-h)
=\lim_{h\to0}\frac{1-cos7(π-h-π)}{5((π-h)-π)^2}
=\lim_{h\to0}\frac{2sin^2(7/2)h}{5h^2}
=\lim_{h\to0}(2/5)(\frac{sin(7/2)h}{(7/2)h})^2×(7/2)^2 = (2/5) × (49/4) = 49/10
Thus, from eq(i) we get,
f(π) = 49/10
Hence, f(x) is continuous at x = π
Question 34. If f(x) = \frac{2x+3sinx}{3x+2sinx} , x ≠ 0 is continuous at x = 0, then find f(0).
Solution:
Given that,
f(x) =
\frac{2x+3sinx}{3x+2sinx} Also, f(x) is continuous at x = 0
So, LHL = RHL = f(0) ......(i)
Let us consider LHL,
\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{2(-h)+3sin(-h)}{3(-h)+2sin(-h)}
=\lim_{h\to0}\frac{-2h-3sinh}{-3h-2sinh}
=\lim_{h\to0}\frac{\frac{2h+3sinh}{h}}{\frac{3h+2sinh}{h}}
=\lim_{h\to0}\frac{2+3\frac{sinh}{h}}{3+2\frac{sinh}{h}}=\frac{2+3}{3+2}=1 From eq(i) we get,
f(0) = 1
Question 35. Find the value of k for which f(x)=\begin{cases}\frac{1-cos4x}{8x^2} ,& \text{when }x\neq0 \\k,& \text{when }x=0\end{cases} is continuous at x = 0
Solution:
Given that,
f(x)=\begin{cases}\frac{1-cos4x}{8x^2} ,& \text{when }x\neq0 \\k,& \text{when }x=0\end{cases} Also, f(x) is continuous at x = 0
LHL = RHL = f(0) .....(i)
f(0) = k
Let us consider LHL,
\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{1-cos4(-h)}{8(-h)^2}
=\lim_{h\to0}\frac{1-cos4h}{8h^2}
=\lim_{h\to0}\frac{2sin^22h}{8h^2}
=\lim_{h\to0}(\frac{sin2h}{2h})^2=1 Thus, from eq(i) we get,
k = 1
Question 36. In each of the following, find the value of the constant k so that the given function is continuous at the indicated point:
(i) f(x)=\begin{cases}\frac{1-cos2kx}{x^2},& \text{if }x\neq0 \\8,& \text{if }x=0\end{cases} at x = 0
Solution:
Given that,
f(x)=\begin{cases}\frac{1-cos2kx}{x^2},& \text{if }x\neq0 \\8,& \text{if }x=0\end{cases} Also, f(x) is continuous at x = 0
\lim_{x\to0}f(x) =f(0) ⇒
\lim_{x\to0}\frac{1-cos2kx}{x^2}=8 ⇒
\lim_{x\to0}\frac{2k^2sin^2kx}{k^2x^2}=8 ⇒
2k^2\lim_{x\to0}(\frac{sinkx}{kx})^2=8 ⇒ 2k2 × 1 = 8
⇒ k2 = 4
⇒ k = ±2
(ii) f(x)=\begin{cases}(x-1)\frac{tanπx}{2},& \text{if }x\neq1 \\k,& \text{if }x=1\end{cases} at x = 1
Solution:
Given that,
f(x)=\begin{cases}(x-1)\frac{tanπx}{2},& \text{if }x\neq1 \\k,& \text{if }x=1\end{cases} Also, f(x) is continuous at x = 1
\lim_{x\to1}f(x) =f(1) ⇒
\lim_{x\to1}(x-1)tan(πx/2)=k Now, on putting x - 1 = y, we get
\lim_{y\to0}ytan\frac{π(y+1)}{2}=k ⇒
\lim_{y\to0}ytan(\frac{πy}{2}+π/2)=k ⇒
\lim_{y\to0}ytan(\frac{π}{2}+\frac{πy}{2})=k ⇒
-\lim_{y\to0}ycot(\frac{π}{2})=k ⇒
\frac{-2}{π}\lim_{y\to0}\frac{\frac{πy}{2}cos(\frac{πy}{2})}{sin\frac{πy}{2}}=k ⇒
\frac{-2}{π}\frac{\lim_{y\to0}cos(\frac{πy}{2})}{\lim_{y\to0}(\frac{sin(\frac{πy}{2})}{\frac{πy}{2}})}=k ⇒ (-2/π) × (1/1) = k
⇒ k = (-2/π)
(iii) f(x)=\begin{cases}k(x^2-2x),& \text{if }x<0 \\cosx,& \text{if }x\geq0\end{cases} at x = 0
Solution:
Given that,
f(x)=\begin{cases}k(x^2-2x),& \text{if }x<0 \\cosx,& \text{if }x\geq0\end{cases} Also, f(x) is continuous at x = 0
Let us consider LHL, at x = 0
\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}f(-h)
=\lim_{h\to0}k(h^2+2h)=0 Let us consider RHL at x = 0
\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)
=\lim_{h\to0}f(h)
=\lim_{h\to0}cosh=1
\lim_{x\to0^-}f(x)≠\lim_{x\to0^+}f(x) Hence, no value of k exists for which function is continuous at x = 0.
(iv) f(x)=\begin{cases}kx+1,& \text{if }x\leqπ \\cosx,& \text{if }x>π\end{cases} at x = π
Solution:
Given that,
f(x)=\begin{cases}kx+1,& \text{if }x\leqπ \\cosx,& \text{if }x>π\end{cases} Also, f(x) is continuous at x = π
Let us consider LHL
\lim_{x\toπ^-}f(x) =\lim_{h\to0}f(π-h)
=\lim_{h\to0}k(π-h)+1=kπ+1 Let us consider RHL
\lim_{x\toπ^+}f(x) =\lim_{h\to0}f(π+h)
=\lim_{h\to0}cos(π+h) cosπ = -1
As we know that f(x) is continuous at x = π, so
\lim_{x\toπ^-}f(x)=\lim_{x\toπ^+}f(x) ⇒ kπ + 1 = -1
⇒ k = (-2/π)
(v) f(x)=\begin{cases}kx+1,& \text{if }x\leq5 \\3x-5,& \text{if }x>5\end{cases} at x = 5
Solution:
Given that,
f(x)=\begin{cases}kx+1,& \text{if }x\leq5 \\3x-5,& \text{if }x>5\end{cases} Also, f(x) is continuous at x = 5
Let us consider LHL
\lim_{x\to5^-}f(x)=\lim_{h\to0}f(5-h)
=\lim_{h\to0}k(5-h)+1 = 5k + 1
Let us consider RHL
\lim_{x\to5^+}f(x)=\lim_{h\to0}f(5+h)
=\lim_{h\to0}3(5+h)-5 = 10
As we know that f(x) is continuous at x = 5, so
\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x) ⇒ 5k + 1 = 10
⇒ k = 9/5
(vi) f(x)=\begin{cases}\frac{x^2-25}{x-5},& \text{if }x\neq5 \\k,& \text{if }x=5\end{cases} at x = 5
Solution:
Given that,
f(x)=\begin{cases}\frac{x^2-25}{x-5},& \text{if }x\neq5 \\k,& \text{if }x=5\end{cases} Also, f(x) is continuous at x = 5
So,
f(x) = (x2 - 25)/(x - 5), if x ≠ 5 & f(x) = k, if x = 5
⇒ f(x)= {(x - 5)(x+5)/(x-5)}, if x ≠ 5 & f(x) = k, if x = 5
⇒ f(x)= (x + 5), if x ≠ 5 & f(x) = k, if x = 5
As we know that f(x) is continuous at x = 5, so
\lim_{x\to5}f(x)=f(5) ⇒
\lim_{x\to5}(x+5)=k ⇒ k = 5 + 5 = 10
(vii) f(x)=\begin{cases}kx^2,& \text{if }x\geq1 \\4,& \text{if }x<1\end{cases} at x = 1
Solution:
Given that,
f(x)=\begin{cases}kx^2,& \text{if }x\geq1 \\4,& \text{if }x<1\end{cases} Also, f(x) is continuous at x = 1
Let us consider LHL
\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}4=4 Let us consider RHL
\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}k(1+h)^2 = k
As we know that f(x) is continuous at x = 1, so
\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x) ⇒ k = 4
(viii) f(x)=\begin{cases}k(x^2+2),& \text{if }x\leq0 \\3x+1,& \text{if }x>0\end{cases} at x = 0
Solution:
Given that,
f(x)=\begin{cases}k(x^2+2),& \text{if }x\leq0 \\3x+1,& \text{if }x>0\end{cases} Also, f(x) is continuous at x = 0
Let us consider LHL
\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}k((-h)^2+2) = 2k
Let us consider RHL
\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}3h+1 = 1
As we know that f(x) is continuous at x = 0, so
\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x) ⇒ 2k = 1
⇒ k = 1/2
(ix) f(x)=\begin{cases}\frac{x^3+x^2-16x+20}{(x-2)^2},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases} at x = 2
Solution:
Given that,
f(x)=\begin{cases}\frac{x^3+x^2-16x+20}{(x-2)^2},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases} Also, f(x) is continuous at x = 2
f(x)=
\frac{x^3+x^2-16x+20}{(x-2)^2} , if x ≠ 2 & f(x) = k, if x = 2⇒ f(x)=
\frac{x^3+x^2-16x+20}{x^2-4x+4} , if x ≠ 2 & f(x) = k, if x = 2⇒ f(x)=
\frac{(x+5)(x^2-4x+4)}{x^2-4x+4} , if x ≠ 2 & f(x) = k, if x = 2⇒ f(x)= (x + 5), if x ≠ 2 & f(x) = k, if x = 2
As we know that f(x) is continuous at x = 2, so
\lim_{x\to2}f(x)=f(2) ⇒
\lim_{x\to2}(x+5)=f(2) ⇒ k = 2 + 5 = 7
Question 37. Find the values of a and b so that the function f given by
f(x)=\begin{cases}1,& \text{if }x\leq3 \\ax+b,& \text{if }3<x<5\\7, &\text{if }x\geq5\end{cases} is continuous at x = 3 and x = 5.
Solution:
Given that,
f(x)=\begin{cases}1,& \text{if }x\leq3 \\ax+b,& \text{if }3<x<5\\7, &\text{if }x\geq5\end{cases} Let us consider LHL at x = 3,
\lim_{x\to3^-}f(x)=\lim_{h\to0}f(3-h)
=\lim_{h\to0}(1) = 1
Let us consider RHL at x = 3,
\lim_{x\to3^+}f(x)=\lim_{h\to0}f(3+h)
=\lim_{h\to0}a(3+h)+b = 3a + b
Let us consider LHL at x = 5,
\lim_{x\to5^-}f(x)=\lim_{h\to0}f(5-h)
=\lim_{h\to0}(a(5-h)+b) = 5a + b
Let us consider RHL at x = 5,
\lim_{x\to5^+}f(x)=\lim_{h\to0}f(5+h)
=\lim_{h\to0}7 = 7
It is given that f(x) is continuous at x = 3 and x = 5, then
\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x) and\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x) ⇒ 1 = 3a + b .....(i)
and 5a + b = 7 .......(ii)
On solving eq(i) and (ii), we get
a = 3 and b = -8
Question 38. If f(x)=\begin{cases}\frac{x^2}{2},& \text{if }0\leq x \leq 1 \\2x^2-3x+(\frac{3}{2}),& \text{if }1<x\leq2\end{cases} . Show that f is continuous at x = 1.
Solution:
Given that,
f(x)=\begin{cases}\frac{x^2}{2},& \text{if }0\leq x \leq 1 \\2x^2-3x+(\frac{3}{2}),& \text{if }1<x\leq2\end{cases} So,
Let us consider LHL at x = 1,
\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}\frac{(1-h)^2}{2} = 1/2
Let us consider RHL at x = 1,
\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}[2(1+h)^2-3(1+h)+3/2] = 2 - 3 + 3/2 = 1/2
Also,
f(1) = (1)2/2 = 1/2
\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1) LHL = RHL = f(1)
Hence, the f(x) is continuous at x = 1
Question 39. Discuss the continuity of the f(x) at the indicated points:
(i) f(x) = |x| + |x - 1| at x = 0, 1.
Solution:
Given that,
f(x) = |x| + |x - 1|
So, here we check the continuity of the given f(x) at x = 0,
Let us consider LHL at x = 0,
\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}[|0-h|+|0-h-1|]=1 Let us consider RHL at x = 0,
\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}[|0+h|+|0+h-1|]=1 Also,
f(0) = |0| + |0 - 1| = 0 + 1 = 1
LHL = RHL = f(0)
Now, we check the continuity of the given f(x) at x = 1,
Let us consider LHL at x = 1,
\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}f(|1-h|+|1-h-1|)=1+0 = 1
Let us consider RHL at x = 1
\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}f(|1+h|+|1+h-1|)=1+0 = 1
Also,
f(1) = |1| + |1 - 1| = 1 + 0 = 1
LHL = RHL = f(1)
Hence, f(x) is continuous at x = 0, 1.
(ii) f(x) = |x - 1| + |x + 1| at x = -1, 1.
Solution:
Given that,
f(x) = |x - 1| + |x + 1| at x = -1, 1.
So, here we check the continuity of the given f(x) at x = -1,
Let us consider LHL at x = -1,
\lim_{x\to-1^-}f(x)=\lim_{h\to0}f(-1-h)
=\lim_{h\to0}[|-1-h-1|+|-1-h+1|]=2+0=2 Let us consider RHL at x = -1,
\lim_{x\to-1^+}f(x)=\lim_{h\to0}f(-1+h)
=\lim_{h\to0}[|-1+h-1|+|-1+h+1|]=2+0=2 Also,
f(-1) = |-1 - 1| + |-1 + 1| = |-2| = 2
LHL = RHL = f(-1)
Now, we check the continuity of the given f(x) at x = 1,
Let us consider LHL at x = 1,
\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}f(|1-h-1|+|1-h+1|)=0+2 = 2
\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}f(|1+h-1|+|1+h+1|)=0+2 = 2
Also,
f(1) = |1 + 1| + |1 - 1| = 2
LHL = RHL = f(1)
Hence, f(x) is continuous at x = -1, 1.
Question 40. Prove that f(x)=\begin{cases}\frac{x-|x|}{x},& \text{if }x\neq0 \\2,& \text{if }x=0\end{cases} is discontinuous at x = 0.
Solution:
Prove that
f(x)=\begin{cases}\frac{x-|x|}{x},& \text{if }x\neq0 \\2,& \text{if }x=0\end{cases} is discontinuous at x = 0.Proof:
Let us consider LHL at x = 0,
\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}f(-h)
=\lim_{h\to0}2=2 Let us consider RHL at x = 0,
\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}f(h)
=\lim_{h\to0}0=0 LHL ≠ RHL
Hence, f(x) is discontinuous at x = 0.
Question 41. If f(x)=\begin{cases}2x^2+k,& \text{if }x\geq0 \\-2x^2+k,& \text{if }x<0\end{cases} then what should be the value of k so that f(x) is continuous at x = 0.
Solution:
Given that,
f(x)=\begin{cases}2x^2+k,& \text{if }x\geq0 \\-2x^2+k,& \text{if }x<0\end{cases} Let us consider LHL at x = 0,
\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}f(-h)
=\lim_{h\to0}-2(-h)^2+k = k
Let us consider RHL at x = 0,
\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}f(h)
=\lim_{h\to0}f(2h^2+k) = k
It is given that f(x) is continuous at x = 0.
LHL = RHL = f(0)
⇒
\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=k k can be any real number.
Question 42. For what value λ of is the function
f(x)=\begin{cases}λ(x^2-2x),& \text{if }x\leq0 \\4x+1,& \text{if }x>0\end{cases} continuous at x = 0 ? What about continuity at x = ±1?
Solution:
Given that,
f(x)=\begin{cases}λ(x^2-2x),& \text{if }x\leq0 \\4x+1,& \text{if }x>0\end{cases} Check for x = 0,
Hence, there is no value of λ for which f(x) is continuous at x = 0.
Now for x = 1,
f(1) = 4x + 1 = 4 × 1 + 1 = 5
Hence, for any values of λ, f is continuous at x = 1.
Now for x = -1,
f(-1) = λ(1 + 2)= 3λ
=\lim_{x\to-1}λ(1+2)=3λ
=\lim_{x\to-1}f(x)=f(-1) Hence, for any values of λ, f is continuous at x=-1.
Question 43. For what values of k is the following function continuous at x = 2?
f(x)=\begin{cases}2x+1,& \text{if }x<2 \\k,& \text{if }x=2\\3x-1,& \text{if }x>2\end{cases}
Solution:
Given that,
f(x)=\begin{cases}2x+1,& \text{if }x<2 \\k,& \text{if }x=2\\3x-1,& \text{if }x>2\end{cases} We have,
Let us consider LHL at x = 2,
=\lim_{x\to2^-}f(x)=\lim_{h\to0}f(2-h)
=\lim_{h\to0}(2(2-h)+1) = 5
Let us consider RHL at x = 2,
\lim_{x\to2^+}f(x)=\lim_{h\to0}f(2+h)
=\lim_{h\to0}3(2+h)-1 = 5
Also,
f(2) = k
It is given that f(x) is continuous at x = 2.
LHL = RHL = f(2)
⇒ 5 = 5 = k
Hence, for k = 5, f(x) is continuous at x = 2.
Question 44. Let f(x)=\begin{cases}1-sin^3x3cos^2x,& \text{if }x<(\frac{π}{2}) \\a,& \text{if }x=(\frac{π}{2})\\\frac{b(1-sinx)}{(π-2x)^2},& \text{if }x>(\frac{π}{2})\end{cases} If f(x) is continuous at x = (π/2), find a and b.
Solution:
Given that,
f(x)=\begin{cases}1-sin^3x3cos^2x,& \text{if }x<(\frac{π}{2}) \\a,& \text{if }x=(\frac{π}{2})\\\frac{b(1-sinx)}{(π-2x)^2},& \text{if }x>(\frac{π}{2})\end{cases} Let us consider LHL at x = π/2
=\lim_{x\to(\frac{π}{2})^-}f(x)=\lim_{h\to0}f(\frac{π}{2}-h)
=\lim_{h\to0}\frac{1-sin^3(\frac{π}{2}-h)}{3cos^2(\frac{π}{2}-h)}
=\lim_{h\to0}\frac{1-cos^3h}{3sin^2h}
=\frac{1}{3}\lim_{h\to0}(\frac{(1-cosh)(1+cos^2h+cosh)}{(1-cosh)(1+cosh)})
=\frac{1}{3}\lim_{h\to0}(\frac{(1+cos^2h+cosh)}{(1+cosh)})
=\frac{1}{3}(\frac{1+1+1}{1+1}) = 1/2
Let us consider RHL at x = π/2
=\lim_{x\to(\frac{π}{2})^+}f(x)=\lim_{h\to0}f(\frac{π}{2}+h)
=\lim_{h\to0}(\frac{b[1-sin(\frac{π}{2}+h)]}{[π-2(\frac{π}{2}+h)]^2})
=\lim_{h\to0}(\frac{b(1-cosh)}{[-2h]^2})
=\lim_{h\to0}(\frac{2bsin^2(\frac{h}{2})}{4h^2})
=\lim_{h\to0}(\frac{2bsin^2(\frac{h}{2})}{\frac{16h^2}{4}})
=(\frac{b}{8})\lim_{h\to0}(\frac{sin(\frac{h}{2})}{\frac{h}{2}})^2 = b/8 × 1
= b/8
Also,
f(π/2) = a
It is given that f(x) is continuous at x = π/2.
LHL = RHL = f(π/2)
So,
⇒ 1/2 = b/8 = a
⇒ a = 1/2 and b = 4
Question 45. If the functions f(x), defined below is continuous at x = 0, find the value of k,
f(x)=\begin{cases}\frac{1-cos2x}{2x^2},& \text{if }x<0 \\k,& \text{if }x=0\\\frac{x}{|x|},& \text{if }x>0\end{cases}
Solution:
Given that,
f(x)=\begin{cases}\frac{1-cos2x}{2x^2},& \text{if }x<0 \\k,& \text{if }x=0\\\frac{x}{|x|},& \text{if }x>0\end{cases} Let us consider LHL at x = 0,
=\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}(\frac{1-cos2(-h)}{2(-h)^2})
=\lim_{h\to0}(\frac{1-cos2h}{2h^2})
=\frac{1}{2}\lim_{h\to0}(\frac{2sin^2h}{h^2})
=\frac{2}{2}\lim_{h\to0}(\frac{sin^2h}{h^2})
=\frac{2}{2}\lim_{h\to0}(\frac{sinh}{h})^2 = 1 × 1
Let us consider RHL at x = 0,
\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}f(h)
=\lim_{h\to0}1=1 Also,
f(0) = k
It is given that f(x) is continuous at x = 0,
LHL = RHL = f(0)
So,
⇒ 1 = 1 = k
Hence, the required value of k is 1.
Question 46. Find the relationship between 'a' and 'b' so that function 'f' defined by
f(x)=\begin{cases}ax+1,& \text{if }x\leq3 \\bx+3,& \text{if }x>3\\\end{cases} is continuous at x = 3.
Solution:
Given that,
f(x)=\begin{cases}ax+1,& \text{if }x\leq3 \\bx+3,& \text{if }x>3\\\end{cases} Let us consider LHL at x = 3,
\lim_{x\to3^-}f(x)=\lim_{h\to0}f(3-h)
=\lim_{h\to0}a(3-h)+1 = 3a + 1
Let us consider RHL at x = 3,
\lim_{x\to3^+}f(x)=\lim_{h\to0}f(3+h)
=\lim_{h\to0}b(3+h)+3 = 3b + 3
It is given that f(x) is continuous at x = 3,
LHL = RHL = f(3)
So,
⇒ 3a + 1 = 3b + 3
⇒ 3a - 3b = 2
Hence, the required relationship between a and b is 3a - 3b = 2.
Summary
This set likely covers more advanced topics such as:
- Continuity of inverse functions
- Continuity and monotonicity
- Discontinuities in rational functions
- Continuity of parametric functions
- Advanced applications of the Intermediate Value Theorem
